
where \( {\mathbf {\Sigma }}_{g_{[k]}}=\frac{1}{\bar{b}}\hat{\bar{{\mathbf {H}}}}_{g_{[k]}}\hat{\bar{{\mathbf {H}}}}^{{\scriptscriptstyle \mathsf {H}}}_{g_{[k]}}+ \alpha {\mathbf {\mathrm {I}}}_{\bar{b}} \) . In (REF ), we applied the matrix inversion lemma twice [1]} and, in (), we used [1]}. The last step makes use of [1]} with \( \bar{m}_{g} \)  and \( \bar{\delta }_{g} \)  defined in Theorem REF . Note that \( \bar{m}_{g} \)  is basically the derivative of \( \bar{\delta }_{g} \)  with respect to \( \alpha \) , i.e., \( \bar{m}_{g}=\bar{\delta }_{g}^{\prime } =\frac{1}{\bar{b}}\left(\bar{{\mathbf {R}}}_{g}{\mathbf {T}}_{g}^{\prime }\right)\) , where \( {\mathbf {T}}_{g}^{\prime }={\mathbf {T}}_{g}\left(\frac{{\bar{K}}}{\bar{b}}\frac{\bar{{\mathbf {R}}}_{g}\bar{\delta }_{g}^{\prime }}{1+\bar{\delta }_{g}}+ {\mathbf {\mathrm {I}}}_{\bar{b}}\right){\mathbf {T}}_{g} \) . Its expression is obtained after some algebraic manipulations.
